AS Physics P1 MCQs Preparation Thread.

[Oliveme]

so is it ok now..???

yes, thank you.

 

[omg]

any1 havin qn ppr and ms of 2000?

 

[applepie1996]

nope the answer is 10 GHz

oh ok sorry

 

[Gémeaux]

Need help with a few question. If someone cud help, please.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Q 8, 25, and 34
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf
Q 27 and 33.

 

[Wanzi21]

hi.. can you help me with this ^^

 

[confused123]

Need help with a few question. If someone cud help, please.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
Q 8, 25, and 34
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
Q 27 and 33.

Hi,
question 8, first you got to find the velocity till 30 meters. then use that initial velocity to find acceleration for last 10 meters.

34) the resistivity increases and voltage increases as well (directly proportional) so gradient has to go more steep as we move along..

2nd ppr: 27th its zero cause force is centripetal, towards the center and the distance moved is not in the direction of force.

33) each resistor originally will share equal power as they have equal resistance (forget the circuit combination for a while)

now remember again: the two resistors Q and R are connected in parallel. so 1/4 + 1/4 = 1/P, this will give us 2W

 

[bloody_mary]

sorry for the late reply... the working i showed first is wrong, I'm sorry if i made you waste your time (found out looking at http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_er.pdf pages 5 and 6)
This is the correct working added a bit more detail, and also d=1/N you can look at this link http://www.s-cool.co.uk/a-level/physics/diffraction/revise-it/diffraction-from-a-diffraction-grating. sorry again

it's ok thanks though

 

[bloody_mary]

PE = KE + WD? What is this formula? :S Thank you so much!

loss of potential energy = gain in kinteic energy + work done by friction

 

[bloody_mary]

hi.. can you help me with this ^^View attachment 12742

frequency = 1/T

so T is 1/50 which is 0.02s which is the same as 20ms so answer is D

 

[Gémeaux]

question 8, first you got to find the velocity till 30 meters. then use that initial velocity to find acceleration for last 10 meters.

sorry? im stuck on this one, cud u please write the solution?

 

[omg]

hv a look at it guys

 

[Wanzi21]

frequency = 1/T

so T is 1/50 which is 0.02s which is the same as 20ms so answer is D

hmm the answer is B, actually

 

[dayeed]

please tell k how to solve q26 and q 36
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf

 

[confused123]

sorry? im stuck on this one, cud u please write the solution?

oh sorry i mixed it up with some other mcq, didn't read all..

s= ut + 1/2 at ^2 , total distance = 80 m , u will be zero, time is 18 seconds,

80= 1/2 into a into 324 , 80/162 = .49 ms^-2

 

[biba]

oh ok sorry

no problem

 

[biba]

obviously they r good! show off!

hahaha i wasn't showing off! i just needed some opinions u know
cuz i sucked in physix my p2 went bad!!!!!

 

[confused123]

hahaha i wasn't showing off! i just needed some opinions u know
cuz i sucked in physix my p2 went bad!!!!!

mission successful, u took it serious, i know i m just having fun.
bdw i didn't sucked that bad in p2. i don't like to ponder over past. shuuu 4get dat ppr. mcqs on my mind now.

 

[biba]

mission successful, u took it serious, i know i m just having fun.
bdw i didn't sucked that bad in p2. i don't like to ponder over past. shuuu 4get dat ppr. mcqs on my mind now.

hahahaha yeah right

 

[waleedsmz]

hv a look at it guys

I believe the answer is C because as the kinetic energy increases, the change in speed required to do this increase actually decreases. This is due to kinetic energy being proportional to the square of the velocity. Let me explain...

Increase in K.E. ----> Increase in speed
1-->10 1-->3.16 (change 2.16)
10-->20 3.16-->4.47 (change 1.31)
20-->30 4.47-->5.48 (change 1.01)

In this example, even though there is a constant increase in K.E., the change in velocity decreases as the K.E. increases.

Hope that helped. : )

 

[omg]

I believe the answer is C because as the kinetic energy increases, the change in speed required to do this increase actually decreases. This is due to kinetic energy being proportional to the square of the velocity. Let me explain...

Increase in K.E. ----> Increase in speed
1-->10 1-->3.16 (change 2.16)
10-->20 3.16-->4.47 (change 1.31)
20-->30 4.47-->5.48 (change 1.01)

In this example, even though there is a constant increase in K.E., the change in velocity decreases as the K.E. increases.

Hope that helped. : )

it sure did
THANKS a lot!
=D